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Question Number 193336 by Abdulz last updated on 10/Jun/23

Commented by Abdulz last updated on 10/Jun/23

$$please,help$$

Answered by Humble last updated on 11/Jun/23

$$$$$$\alpha =45°,\theta =30°v_0=4m/s$$$$v_2=?g=9.8m/s$$$$H_{max}=\frac{u^2}{2a}=\frac{v_0^2{cos}^2\left(\theta \right)}{2gsin\left(\alpha \right)}$$$$=\frac{4^2{cos}^2\left(30\right)}{2\times 9.8\times sin\left(45\right)}$$$$=\frac{16\times {\left(\frac{\sqrt{3}}{2}\right)}^2}{19.6\times \frac{\sqrt{2}}{2}}=\frac{16\times \frac{3}{4}}{\frac{19.6\sqrt{2}}{2}}$$$$=16\times \frac{3}{4}\times \frac{2}{19.6\sqrt{2}}$$$$=16\times \frac{3}{2}\times \frac{1}{19.6\sqrt{2}}$$$$=\frac{24}{19.6\sqrt{2}}=\frac{30\sqrt{2}}{49}=0.87m$$$$\left(b\right)R=\frac{2v_0^{2}cos\left(\theta \right)sin\left(\theta \right)}{g}$$$$R=\frac{2{\left(4\right)}^{2}cos\left(30\right)sin\left(30\right)}{9.8}$$$$R=\frac{32\times \frac{\sqrt{3}}{2}\times \frac{1}{2}}{9.8}=\frac{\frac{32\sqrt{3}}{4}}{9.8}$$$$R=\frac{32\sqrt{3}}{4}\times \frac{1}{9.8}=\frac{32\sqrt{3}}{39.2}$$$$R=\frac{40\sqrt{3}}{49}=1.4m$$

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